Let the balloon rise

Problem descriptioncontest time again! How excited it is to see balloons floating around. but to tell you a secret, the judges 'favorite time is guessing the most popular problem. when the contest is over, they will count the balloons of each color and find the result.

This year, they decide to leave this lovely job to you.

Inputinput contains multiple test cases. each test case starts with a number N (0 <n <= 1000) -- The total number of balloons distributed. the next n lines contain one color each. the color of a balloon is a string of up to 15 lower-case letters.

A test case with n = 0 terminates the input and this test case is not to be processed.

 

Outputfor each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

Sample Input

5greenredblueredred3pinkorangepink0

 

Sample output

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int maxindex(int *p,int n)
{
int i;
int tem;
int index=0;
    tem=p[0];
for(i=0;i<n;i++)
    {
if(tem<=p[i])
            tem=p[i];
    }
for(i=0;i<n;i++)
    {
if(tem==p[i])
        {
            index=i;
        }
    }
return index;
}
void main()
{
int n;
int i=0;
int j=0;
char *p[1000];
int level=0;
char *save[1000];

int cal[1000]={0};
int index=-1;
for(i=0;i<1000;i++)
    {
        save[i]=(char*)malloc(sizeof(char)*20); 
    }
while(scanf("%d",&n)&&n!=0)
    {  
    index++;

for(i=0;i<n;i++)
    {
        p[i]=(char*)malloc(sizeof(char)*20);
        scanf("%s",p[i]);      
    }
for(i=0;i<n;i++)
        cal[i]=0;

for(level=0;level<n;level++)
    {
for(j=level+1;j<n;j++)
if(!strcmp(p[level],p[j]))
                cal[level]++;

    }
    strcpy(save[index],p[maxindex(cal,n)]);
for(i=0;i<n;i++)
    {
    free(p[i]);
    p[i]=NULL;
    }
    }

for(i=0;i<=index;i++)
        printf("%s\n",save[i]);



}
red
pink
 
AuthorWU, Jiazhi 
SourceZJCPC2004